3.2202 \(\int \sqrt{a+b x} (A+B x) \sqrt{d+e x} \, dx\)

Optimal. Leaf size=196 \[ \frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (2 A b e-B (a e+b d))}{8 b^2 e^2}-\frac{(b d-a e)^2 (2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{5/2} e^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{d+e x} (2 A b e-B (a e+b d))}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e} \]

[Out]

((b*d - a*e)*(2*A*b*e - B*(b*d + a*e))*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^2*e^2) + ((2*A*b*e - B*(b*d + a*e))*(
a + b*x)^(3/2)*Sqrt[d + e*x])/(4*b^2*e) + (B*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b*e) - ((b*d - a*e)^2*(2*A*b*
e - B*(b*d + a*e))*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(5/2)*e^(5/2))

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Rubi [A]  time = 0.150053, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {80, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (2 A b e-B (a e+b d))}{8 b^2 e^2}-\frac{(b d-a e)^2 (2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{5/2} e^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{d+e x} (2 A b e-B (a e+b d))}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(A + B*x)*Sqrt[d + e*x],x]

[Out]

((b*d - a*e)*(2*A*b*e - B*(b*d + a*e))*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^2*e^2) + ((2*A*b*e - B*(b*d + a*e))*(
a + b*x)^(3/2)*Sqrt[d + e*x])/(4*b^2*e) + (B*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b*e) - ((b*d - a*e)^2*(2*A*b*
e - B*(b*d + a*e))*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(5/2)*e^(5/2))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+b x} (A+B x) \sqrt{d+e x} \, dx &=\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}+\frac{\left (3 A b e-B \left (\frac{3 b d}{2}+\frac{3 a e}{2}\right )\right ) \int \sqrt{a+b x} \sqrt{d+e x} \, dx}{3 b e}\\ &=\frac{(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt{d+e x}}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}+\frac{\left ((b d-a e) \left (3 A b e-B \left (\frac{3 b d}{2}+\frac{3 a e}{2}\right )\right )\right ) \int \frac{\sqrt{a+b x}}{\sqrt{d+e x}} \, dx}{12 b^2 e}\\ &=\frac{(b d-a e) (2 A b e-B (b d+a e)) \sqrt{a+b x} \sqrt{d+e x}}{8 b^2 e^2}+\frac{(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt{d+e x}}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac{\left ((b d-a e)^2 \left (3 A b e-B \left (\frac{3 b d}{2}+\frac{3 a e}{2}\right )\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{24 b^2 e^2}\\ &=\frac{(b d-a e) (2 A b e-B (b d+a e)) \sqrt{a+b x} \sqrt{d+e x}}{8 b^2 e^2}+\frac{(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt{d+e x}}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac{\left ((b d-a e)^2 \left (3 A b e-B \left (\frac{3 b d}{2}+\frac{3 a e}{2}\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{12 b^3 e^2}\\ &=\frac{(b d-a e) (2 A b e-B (b d+a e)) \sqrt{a+b x} \sqrt{d+e x}}{8 b^2 e^2}+\frac{(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt{d+e x}}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac{\left ((b d-a e)^2 \left (3 A b e-B \left (\frac{3 b d}{2}+\frac{3 a e}{2}\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{12 b^3 e^2}\\ &=\frac{(b d-a e) (2 A b e-B (b d+a e)) \sqrt{a+b x} \sqrt{d+e x}}{8 b^2 e^2}+\frac{(2 A b e-B (b d+a e)) (a+b x)^{3/2} \sqrt{d+e x}}{4 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{3/2}}{3 b e}-\frac{(b d-a e)^2 (2 A b e-B (b d+a e)) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{5/2} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.10943, size = 201, normalized size = 1.03 \[ \frac{(a+b x)^{3/2} (d+e x)^{3/2} \left (3 B-\frac{9 (a B e-2 A b e+b B d) \left (\sqrt{e} (a+b x) \sqrt{b d-a e} \sqrt{\frac{b (d+e x)}{b d-a e}} (a e+b (d+2 e x))-\sqrt{a+b x} (b d-a e)^2 \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )\right )}{8 e^{3/2} (a+b x)^2 (b d-a e)^{3/2} \left (\frac{b (d+e x)}{b d-a e}\right )^{3/2}}\right )}{9 b e} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]*(A + B*x)*Sqrt[d + e*x],x]

[Out]

((a + b*x)^(3/2)*(d + e*x)^(3/2)*(3*B - (9*(b*B*d - 2*A*b*e + a*B*e)*(Sqrt[e]*Sqrt[b*d - a*e]*(a + b*x)*Sqrt[(
b*(d + e*x))/(b*d - a*e)]*(a*e + b*(d + 2*e*x)) - (b*d - a*e)^2*Sqrt[a + b*x]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/
Sqrt[b*d - a*e]]))/(8*e^(3/2)*(b*d - a*e)^(3/2)*(a + b*x)^2*((b*(d + e*x))/(b*d - a*e))^(3/2))))/(9*b*e)

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Maple [B]  time = 0.013, size = 755, normalized size = 3.9 \begin{align*} -{\frac{1}{48\,{b}^{2}{e}^{2}}\sqrt{bx+a}\sqrt{ex+d} \left ( -16\,B{x}^{2}{b}^{2}{e}^{2}\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+6\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}b{e}^{3}-12\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{2}d{e}^{2}+6\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{2}e-24\,A\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}x{b}^{2}{e}^{2}-3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{3}{e}^{3}+3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}bd{e}^{2}+3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{2}{d}^{2}e-3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{3}-4\,B\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}xab{e}^{2}-4\,B\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}x{b}^{2}de-12\,A\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}ab{e}^{2}-12\,A\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}{b}^{2}de+6\,B\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}{a}^{2}{e}^{2}-4\,B\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}abde+6\,B\sqrt{be{x}^{2}+aex+bdx+ad}\sqrt{be}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{be{x}^{2}+aex+bdx+ad}}}{\frac{1}{\sqrt{be}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)*(e*x+d)^(1/2),x)

[Out]

-1/48*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(-16*B*x^2*b^2*e^2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+6*A*ln(1/2*(2
*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*e^3-12*A*ln(1/2*(2*b*x*e+2*(b
*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e^2+6*A*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e
*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d^2*e-24*A*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/
2)*x*b^2*e^2-3*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*e^3+3
*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d*e^2+3*B*ln(1/2*
(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d^2*e-3*B*ln(1/2*(2*b*x*e+2
*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d^3-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)
*(b*e)^(1/2)*x*a*b*e^2-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*x*b^2*d*e-12*A*(b*e*x^2+a*e*x+b*d*x+a*d
)^(1/2)*(b*e)^(1/2)*a*b*e^2-12*A*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*b^2*d*e+6*B*(b*e*x^2+a*e*x+b*d*x+
a*d)^(1/2)*(b*e)^(1/2)*a^2*e^2-4*B*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)*a*b*d*e+6*B*(b*e*x^2+a*e*x+b*d*
x+a*d)^(1/2)*(b*e)^(1/2)*b^2*d^2)/(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)/b^2/e^2/(b*e)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)*(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.22334, size = 1168, normalized size = 5.96 \begin{align*} \left [-\frac{3 \,{\left (B b^{3} d^{3} -{\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} e -{\left (B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} +{\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3}\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} - 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \,{\left (8 \, B b^{3} e^{3} x^{2} - 3 \, B b^{3} d^{2} e + 2 \,{\left (B a b^{2} + 3 \, A b^{3}\right )} d e^{2} - 3 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3} + 2 \,{\left (B b^{3} d e^{2} +{\left (B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{96 \, b^{3} e^{3}}, -\frac{3 \,{\left (B b^{3} d^{3} -{\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} e -{\left (B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} +{\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3}\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (8 \, B b^{3} e^{3} x^{2} - 3 \, B b^{3} d^{2} e + 2 \,{\left (B a b^{2} + 3 \, A b^{3}\right )} d e^{2} - 3 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3} + 2 \,{\left (B b^{3} d e^{2} +{\left (B a b^{2} + 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{48 \, b^{3} e^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)*(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(B*b^3*d^3 - (B*a*b^2 + 2*A*b^3)*d^2*e - (B*a^2*b - 4*A*a*b^2)*d*e^2 + (B*a^3 - 2*A*a^2*b)*e^3)*sqrt
(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 - 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt
(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(8*B*b^3*e^3*x^2 - 3*B*b^3*d^2*e + 2*(B*a*b^2 + 3*A*b^3)*d*e^2 - 3*(B
*a^2*b - 2*A*a*b^2)*e^3 + 2*(B*b^3*d*e^2 + (B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^3),
-1/48*(3*(B*b^3*d^3 - (B*a*b^2 + 2*A*b^3)*d^2*e - (B*a^2*b - 4*A*a*b^2)*d*e^2 + (B*a^3 - 2*A*a^2*b)*e^3)*sqrt
(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d
*e + a*b*e^2)*x)) - 2*(8*B*b^3*e^3*x^2 - 3*B*b^3*d^2*e + 2*(B*a*b^2 + 3*A*b^3)*d*e^2 - 3*(B*a^2*b - 2*A*a*b^2)
*e^3 + 2*(B*b^3*d*e^2 + (B*a*b^2 + 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B x\right ) \sqrt{a + b x} \sqrt{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)**(1/2)*(e*x+d)**(1/2),x)

[Out]

Integral((A + B*x)*sqrt(a + b*x)*sqrt(d + e*x), x)

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Giac [A]  time = 1.41008, size = 439, normalized size = 2.24 \begin{align*} \frac{\frac{20 \,{\left (\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} e^{\left (-2\right )}}{b^{4}} + \frac{{\left (b d e - a e^{2}\right )} e^{\left (-4\right )}}{b^{4}}\right )} + \frac{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} e^{\left (-\frac{7}{2}\right )} \log \left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac{7}{2}}}\right )} A{\left | b \right |}}{b^{2}} + \frac{{\left (\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} e^{\left (-2\right )}}{b^{6}} + \frac{{\left (b d e^{3} - 7 \, a e^{4}\right )} e^{\left (-6\right )}}{b^{6}}\right )} - \frac{3 \,{\left (b^{2} d^{2} e^{2} - a^{2} e^{4}\right )} e^{\left (-6\right )}}{b^{6}}\right )} - \frac{3 \,{\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} e^{\left (-\frac{9}{2}\right )} \log \left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac{11}{2}}}\right )} B{\left | b \right |}}{b^{3}}}{1920 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)*(e*x+d)^(1/2),x, algorithm="giac")

[Out]

1/1920*(20*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e^(-2)/b^4 + (b*d*e - a*e^2)*e^(-4)
/b^4) + (b^2*d^2 - 2*a*b*d*e + a^2*e^2)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x +
a)*b*e - a*b*e)))/b^(7/2))*A*abs(b)/b^2 + (sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(
b*x + a)*e^(-2)/b^6 + (b*d*e^3 - 7*a*e^4)*e^(-6)/b^6) - 3*(b^2*d^2*e^2 - a^2*e^4)*e^(-6)/b^6) - 3*(b^3*d^3 - a
*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*e^(-9/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b
*e - a*b*e)))/b^(11/2))*B*abs(b)/b^3)/b